Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $a \neq 0$. $z = \dfrac{-2a + 14}{a^2 - 12a + 35} \div \dfrac{a + 1}{-a^2 - 6a - 5} $
Explanation: Dividing by an expression is the same as multiplying by its inverse. $z = \dfrac{-2a + 14}{a^2 - 12a + 35} \times \dfrac{-a^2 - 6a - 5}{a + 1} $ First factor out any common factors. $z = \dfrac{-2(a - 7)}{a^2 - 12a + 35} \times \dfrac{-(a^2 + 6a + 5)}{a + 1} $ Then factor the quadratic expressions. $z = \dfrac {-2(a - 7)} {(a - 7)(a - 5)} \times \dfrac {-(a + 1)(a + 5)} {a + 1} $ Then multiply the two numerators and multiply the two denominators. $z = \dfrac {-2(a - 7) \times -(a + 1)(a + 5) } { (a - 7)(a - 5) \times (a + 1)} $ $z = \dfrac {2(a + 1)(a + 5)(a - 7)} {(a - 7)(a - 5)(a + 1)} $ Notice that $(a - 7)$ and $(a + 1)$ appear in both the numerator and denominator so we can cancel them. $z = \dfrac {2(a + 1)(a + 5)\cancel{(a - 7)}} {\cancel{(a - 7)}(a - 5)(a + 1)} $ We are dividing by $a - 7$ , so $a - 7 \neq 0$ Therefore, $a \neq 7$ $z = \dfrac {2\cancel{(a + 1)}(a + 5)\cancel{(a - 7)}} {\cancel{(a - 7)}(a - 5)\cancel{(a + 1)}} $ We are dividing by $a + 1$ , so $a + 1 \neq 0$ Therefore, $a \neq -1$ $z = \dfrac {2(a + 5)} {a - 5} $ $ z = \dfrac{2(a + 5)}{a - 5}; a \neq 7; a \neq -1 $